What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

-30.02 ºC

Explanation:

Assuming the antifreeze to be ethylene glycol (C₂H₆O₂) which is popular antifreeze.

Molar mass of ethylene glycol (C₂H₆O₂) = 62g/mol

Step 1: calculate the freezing point depression of the solution

ΔT = -Kf*M

where,

ΔT= depression in the freezing point.

M = the molarity of the solution (mol solute / Kg solvent)

Kf = molar freezing point constant  of water = 1.86°C/m

To determine depression in the freezing point (ΔT), first we need to calculate;

Step 2: calculate the molarity of the solute (ethylene glycol)

Molar mass ethylene glycol = 62 g/mol

molarity of ethylene glycol in mol = 50 g / 62g/mol = 0.807 mol

Step 3: calculate mass of solvent in kg

There is 1kg of ethylene glycol which is present in 1kg of water

mass of solvent (water) in kg= 50 g/ 1000 g/ Kg = 0.050 Kg

Step 4: calculate the molarity of the solution (M)

M = 0.807 mol / 0.050 Kg = 16.14 m

Step 5: calculate the freezing point depression of the solution  (ΔT)

ΔT = - Kf*M = -1.86 ºC/m x 16.14 m

     = -30.02 ºC

the answer would be c. motion and direction

i hope this : )

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